Question

Calculate 1/(1*3) + 1/(2*4) + 1/(3*5) + ... + 1/(n*(n+2))

Answer 1

`$\frac 3 4$`

Explanation:

We have the following series

`$\sum_(i=1)^(\infty){(1)/(n(n+2)) } = (1)/(1\cdot 3)+(1)/(2\cdot 4)+(1)/(3\cdot 5)+\dots $`

From this series, you can note a telescope series, thus

`$\sum_(n=1)^\infty (1)/(n(n+2))=\frac1 2\sum_(n=1)^\infty\left(\frac 1 n-\frac 1{n+2}\right)$`

Now we just have to calculate the limit as
`n\to\infty`

For this case we can see that

`$\sum_(n=1)^m\left(\frac1 n-\frac1{n+2}\right)=1+\frac12-\frac1{m+1}-\frac1{m+2}$`

This is because there is a pattern

`$\left((1)/(1){-(1)/(3)}\right)+\left((1)/(2){-(1)/(4)}\right)+\left((1)/(3)-(1)/(5)\right)+\left((1)/(4)-(1)/(6)\right)+\left((1)/(5)-(1)/(7)\right)+\left((1)/(6)-(1)/(8)\right)+\dots}$`

in which every fraction from
`$(1)/(3) $` get cancelled, it implies that we are only left with

`$(1)/(1) +(1)/(2) = 1+(1)/(2) =(3)/(2) $`

Therefore,

`$\sum_(n=1)^\infty (1)/(n(n+2))=\frac1 2\sum_(n=1)^\infty\left(\frac 1 n-\frac 1{n+2}\right) = \frac1 2 \left(\frac 3 2\right) = \boxed{\frac 3 4}$`